If the distance between the plates of a parallel plate `200 mu F` capacitor charged to 400 V is halved. Calculate the change in energy stored in it.

A capacitor with a capacitane of 50muF is connected to 200V, if the distance between its plates is halved, what will be the potential between the plates and what will be the change in energy stored in it?

A parallel plate capacitor is charged. If the plates are pulled apart

The force between the plates of a parallel plates capacitor of area A, capacitance C and having charge of will be

Two parallel plate capacitors A and B have the same separation d= 8.85xx10^-4m between the plates. The plate area of A and B are 0.04m^2 and 0.02m^2 respectively. A slab of dielectric constant (relative permittivity) K=9 has dimensions such that it can exactly fill the space between the plates of capacitor B. A is then charged to a potential difference of 110V. Calculate the capacitance of A and the energy stored in it.

Derive an expression for energy density of a parallel plate capacitor.

A capacitor of capacitance 50 muF is charged to a potential of 1,000 V. Calculate the energy stored in the capacitor.

What is the area of the plates of a 3 F parallel plate capacitor, if the separation between the plates is 5 mm?

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an isulating handle. As a result the potential difference between the plates

A parallel plate capacitor is charged to a potential difference V by a d.c. source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following will change. energy stored in the capacitor.